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View Full Version : Another Logiz mathematic ?


ignatiuswong
17th March 2008, 06:59 AM
Dear All,

Could you solve it ?

A + B = C
A x B = C

But A must not same B, and The result was same

Example :

2 + 2 = 4
2 x 2 = 4 , it was break the rules either A and B was same values,

Hope you enjoy the games,

Please mention it A,B and C ?

Best Regards,:biglaugh:

Tim Folkerts
17th March 2008, 01:04 PM
Here are a few possibilities ....

A B C
1.382 3.618 5
1.268 4.732 6
1.209 5.791 7
1.172 6.828 8
1.146 7.854 9
1.127 8.873 10
1.113 9.887 11
1.101 10.899 12

Tim Folkerts
17th March 2008, 01:17 PM
With negative numbers values of C, there are more possibilites ...

* One of those possibilities involves nice simple numbers!
* Another possibilitie is related to the "golden ratio".
A = -1.618...
B = 0.618...
C = -1


Tim

ignatiuswong
17th March 2008, 10:16 PM
With negative numbers values of C, there are more possibilites ...

* One of those possibilities involves nice simple numbers!
* Another possibilitie is related to the "golden ratio".
A = -1.618...
B = 0.618...
C = -1


Tim

Tim, if you evaluating the result there still have a decimal, it means break the rules again, concern about your answered above,
it it multiply between A and B the result shown 0.7218 but iif it sum, the result shown -1

maybe you could evaluate this answered

A = 1.5
B = 3

C= 4.5

or A = -1
B = 0.5
C = -0.5

Thanks and Best regards,
:biglaugh:

RLewing
18th March 2008, 06:45 AM
A + B = A * B
A + B - A * B = 0
A + B * (1 - A) = 0
B * (1 - A) = -A

So all solutions where B = A / (A - 1) satisfy the formula.

If you add the requirement of these being integers, then A = 0 and A = 2 are the only ones where A/(A-1) produce integer B (all others are fractions/decimals), but then your requirement of "NOT A=B" is not satisfied.

Unless you choose the A = 1 and divide by zero and get B = infinity. But I am not sure if infinity is an integer.

Infinity + anything = infinity * anything

Dean Frederickson
18th March 2008, 08:40 AM
:topic: a little bit. Did anyone out there know you can take any two digit number, say for example 54 you add the 5 and 4 it equals 9 then subtract 9 from 54 that equals 45 then you add 4 plus 5 and it equals 9. That will work with any two digit number.
72 7+2=9 72-9=63 6+3=9
99 9+9=18 99-18=81 8+1=9
Just thought it was interesting, hope some one else does too.:cool:

Tim Folkerts
18th March 2008, 11:35 AM
Tim, if you evaluating the result there still have a decimal, it means break the rules again, concern about your answered above,
it it multiply between A and B the result shown 0.7218 but iif it sum, the result shown -1
:biglaugh:


I think you multiplied by -1.168, not -1.618. With -1.618, the product is indeed -1.

I hadn't found 1.5+3 = 1.5 * 3 = 4.5 -- another simple solution.


I did about what RLewing did, except I found A & B in terms of C, rather then B & C in terms of A. So I can find the solutions where C is a particular number, whereas he can make A a particular number.
I actually like his approach better. My solution ended up with a quadratic equation to solve.


Tim F

Arjay
3rd July 2008, 06:30 PM
:topic: a little bit. Did anyone out there know you can take any two digit number, say for example 54 you add the 5 and 4 it equals 9 then subtract 9 from 54 that equals 45 then you add 4 plus 5 and it equals 9. That will work with any two digit number.
72 7+2=9 72-9=63 6+3=9
99 9+9=18 99-18=81 8+1=9
Just thought it was interesting, hope some one else does too.:cool:

This fact is actually one of the basic rules for easily checking for divisibility, in this case, checking for divisibility by 9. So if this interests you then you might want to search for those other rules too. Sorry, I'm too lazy to search for them myself right now so I can just hand them over to you right here. :bonk: